In how many ways can '$6$' things be distributed equally among $2$ groups ?

I tried

$\dbinom{6}{3}\times \dbinom{3}{3}\times 3!\times 3!$

But I am not sure if it is correct .

I look for a short and simple way.

I have studied maths up to $12$th grade.

  • $\begingroup$ Why have you multiplied by 3! and 3! ? $\endgroup$ – Shailesh Sep 12 '15 at 1:25
  • $\begingroup$ To permute $3$ objects. $\endgroup$ – R K Sep 12 '15 at 1:28
  • $\begingroup$ Ncmathsadist has already answered the question. $\endgroup$ – Shailesh Sep 12 '15 at 1:29
  • $\begingroup$ Why scare quotes around 6? Is is not really 6 after all? $\endgroup$ – Henning Makholm Sep 12 '15 at 1:30
  • $\begingroup$ actually it was $'3\times 2'$. $\endgroup$ – R K Sep 12 '15 at 1:31

It's $6\choose 3$ if you have groups $A$ and $B$. You choose three items and give them to $A$; there are $6\choose 3$ ways to do this. You then give the remaining to group $B$.

The $3!s$ are called for if you stipulate the order in which the groups receive the items is important. Otherwise, not.

$${6\choose 3} = 20.$$

If the two groups are indistinguishable and you are just interested in partitioning the items into two parts of size three, we have an overcount of a factor of 2. In that case there are only $10$ ways. You must be careful about exactly what you are counting.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.